Question

One kg of air, initially at a temperature of $127\text{ }^{\circ}\text{C}$ expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is $287\text{ J/kg K}$, the magnitude of work transfer is _______ kJ.

• A. 168.12
• B. 203.48
• C. 196.56
• D. 114.8

Hint:

For any ideal gas isobaric expansion where volume is doubled, the temperature also doubles in Kelvin.
Thus, the temperature change ($\Delta T$) is exactly equal to the initial absolute temperature $T_1$, simplifying the work formula to $W = m R T_1$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
This question requires us to compute the boundary work done by an ideal gas (air) undergoing a reversible isobaric (constant pressure) expansion.

Step 2: Key Formula or Approach:
The work done during a constant-pressure process is given by:
\[ W = \int P dV = P(V_2 - V_1) \]
Using the ideal gas equation of state $PV = mRT$, we can rewrite this as:
\[ W = m R (T_2 - T_1) \]
For a constant-pressure process:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Step 3: Detailed Explanation:

• Identify the given parameters:
- Mass of air, $m = 1 \text{ kg}$
- Initial temperature, $T_1 = 127\text{ }^{\circ}\text{C} = 127 + 273 = 400 \text{ K}$
- Final volume is doubled, so $V_2 = 2 V_1$
- Gas constant of air, $R = 287 \text{ J/kg K}$

• Find the final temperature $T_2$ using Charles's law:
\[ T_2 = T_1 \left(\frac{V_2}{V_1}\right) = 400 \times 2 = 800 \text{ K} \]

• Calculate the work done:
\[ W = m \cdot R \cdot (T_2 - T_1) \]
\[ W = 1 \text{ kg} \times 287 \text{ J/kg K} \times (800 - 400) \text{ K} \]
\[ W = 287 \times 400 = 114800 \text{ J} \]

• Convert the work to kilojoules (kJ):
\[ W = \frac{114800}{1000} = 114.8 \text{ kJ} \]

Step 4: Final Answer:
The magnitude of work transfer is $114.8 \text{ kJ}$.